Stoichiometry: finding the yield in reaction of sodium azide with iron oxide

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SUMMARY

The discussion focuses on calculating the yield of nitrogen gas from the reaction between sodium azide (NaN3) and iron oxide (Fe2O3). Given 390 g of sodium azide and 400 g of iron oxide, the yield of metallic iron produced is approximately 89.2%. The volume of nitrogen gas produced under specified conditions (0.82 atm and 27°C) is calculated to be approximately 240.82 L. The calculations utilize the ideal gas law (PV = nRT) and stoichiometric relationships to determine limiting reactants and final outputs.

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Thank you in advance.

Homework Statement



Nitrogen gas can be obtained from the reaction between sodium azide and iron oxide.

6NaN_3_{(s)}+Fe_2O_3_{(s)} \rightarrow 3Na_2O_{(s)}+2Fe_{(s)}+9N_2_{(g)}

a) What is the yield of this reaction, knowing that, from 390 g of azide and 400 g of iron oxide, 100 g of metalic iron were produced?

b) What is the volume (in litres = L) of N2 produced, in a pression of 0.82 atm and a temperature of 27oC, in the conditions given in item a?

Universal constant of gases = R = 0.082 (atm L)/(mol K).

Homework Equations



PV = nRT (P = atm, V = litres, n = mol, R = universal constant of gases, T = Kelvin).

The Attempt at a Solution



a) First, the molar relation between sodium azide and iron oxide, in order to discover what is the limiting reactant:

Mass of 6NaN3 = 6(23 + 3(16)) = 6(23 + 48) = 6(65) = 390 g.
Mass of Fe2O3 = 2(56) + 3(16) = 112 + 48 = 160 g.
Mass of 2Fe = 2(56) = 112 g.
6 mol NaN_3 \rightarrow 1 mol Fe_2O_3
390 g NaN_3\rightarrow 160 g Fe_2O_3

Since we already have 390 g of NaN3, a rule of three will not be necessary.
This mass of sodium azide produces 160 g of iron oxide; thus, there is excess of iron oxide. Therefore, the limiting reactant is NaN3.

Mass of metalic iron produced:
6 mol NaN_3 \rightarrow 2 mol Fe
390 g \rightarrow 112 g
The mass of metalic iron is 112g. Since the problem states the mass produced is 100 g, dividing 100 / 112 will give the yield (R):
R = \frac{100}{112}
R equals approximately 89.2%.

b) Convert 27oC to Kelvin = 27 + 273 = 300 K.
To find the molar volume (Vm): PV_m = nRT, then 0.82V_m = 0.082 \times 300; thus Vm = 30 L.
Since 390 g of NaN3 equal 6 mol, then the volume of N2 produced is 9Vm.
V = 9V_m = 9 \times 30 = 270 L
But the yield is 89.2%:
V = 0.892 \times 270
V equals 240.82 L.
 
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a OK, for b I got a little bit more. Not much. Don't round down intermediate results.
 
Borek said:
a OK, for b I got a little bit more. Not much. Don't round down intermediate results.

Thank you for the response. This calculation was done by hand (100/112 = 50/58 = 25/28 = ~89,2). The next digit would be 8, which makes difference in the result (270 * 89.28 = 271.056). Anyway, if we were to do the calculation of the volume like this:
V = \frac{100 \times 270}{112} = \frac{25 * 270}{28} = \frac{25 * 135}{14}
the result would be better.
It is very good to know that the reasoning to solve the problem is right.
 
It is possible that part of the difference is in molar masses, I am using my stoichiometry calculator and it does all calculations using as exact molar masses as possible.
 

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